Rate of effusion of LPG (a mixture of `n`-butane and propane) is 1.25 times that of `SO_(3)`. Hence, mole fraction of `n`-butane in LPG is

Options

A. 0.75
B. 0.25
C. 0.5
D. 0.67

Explanation

To find the mole fraction of `n`-butane in LPG (a mixture of `n`-butane and propane), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between the rates of effusion and molar mass**:
The rate of effusion of a gas is inversely proportional to the square root of its molar mass. This can be expressed mathematically as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \(R_1\) and \(R_2\) are the rates of effusion of the two gases, and \(M_1\) and \(M_2\) are their molar masses.

2. **Set up the equation for the given problem**:
We know that the rate of effusion of LPG (which we will denote as \(R_{LPG}\)) is 1.25 times that of \(SO_3\) (denote as \(R_{SO_3}\)). Therefore:
\[ R_{LPG} = 1.25 \times R_{SO_3} \]
Using the relationship from step 1, we can write:
\[ \frac{R_{LPG}}{R_{SO_3}} = \sqrt{\frac{M_{SO_3}}{M_{LPG}}} \]

3. **Calculate the molar mass of \(SO_3\)**:
The molar mass of \(SO_3\) can be calculated as:
\[ M_{SO_3} = 32 \, (\text{for S}) + 3 \times 16 \, (\text{for O}) = 32 + 48 = 80 \, \text{g/mol} \]

4. **Substitute the known values into the equation**:
Substituting the values we have:
\[ 1.25 = \sqrt{\frac{80}{M_{LPG}}} \]

5. **Square both sides to eliminate the square root**:
\[ (1.25)^2 = \frac{80}{M_{LPG}} \]
\[ 1.5625 = \frac{80}{M_{LPG}} \]

6. **Rearrange to solve for \(M_{LPG}\)**:
\[ M_{LPG} = \frac{80}{1.5625} \approx 51.2 \, \text{g/mol} \]

7. **Set up the equation for the mean molar mass of the mixture**:
The mean molar mass of a mixture can be expressed as:
\[ M_{LPG} = X_{n-butane} \cdot M_{n-butane} + (1 - X_{n-butane}) \cdot M_{propane} \]
where \(X_{n-butane}\) is the mole fraction of `n`-butane, \(M_{n-butane} = 58 \, \text{g/mol}\), and \(M_{propane} = 44 \, \text{g/mol}\).

8. **Substituting the known values**:
\[ 51.2 = X_{n-butane} \cdot 58 + (1 - X_{n-butane}) \cdot 44 \]

9. **Expand and rearrange the equation**:
\[ 51.2 = 58X_{n-butane} + 44 - 44X_{n-butane} \]
\[ 51.2 = 14X_{n-butane} + 44 \]
\[ 51.2 - 44 = 14X_{n-butane} \]
\[ 7.2 = 14X_{n-butane} \]

10. **Solve for \(X_{n-butane}\)**:
\[ X_{n-butane} = \frac{7.2}{14} = 0.5143 \]

### Final Answer:
The mole fraction of `n`-butane in LPG is approximately **0.5143**.